Simplify the following expression and state the condition under which the simplification is valid: $z = \dfrac{k^2 + 13k + 42}{k^2 + 7k}$
Solution: First factor the expressions in the numerator and denominator. $ \dfrac{k^2 + 13k + 42}{k^2 + 7k} = \dfrac{(k + 6)(k + 7)}{(k)(k + 7)} $ Notice that the term $(k + 7)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(k + 7)$ gives: $z = \dfrac{k + 6}{k}$ Since we divided by $(k + 7)$, $k \neq -7$. $z = \dfrac{k + 6}{k}; \space k \neq -7$